Two parallel long wire carry currents `i_(1)` and `i_(2)` with `i_(1) gt i_(2)` . When the currents are in the same direction , the magnetic field midway between the wires is `10 muT` . When the direction of `i_(2)` is reversed, it becomes `40 muT` . The ratio `i_(1)//i_(2)` is

Options

A. `3:4`
B. `11:7`
C. `7:11`
D. `5:3`

Answer

To solve the problem step by step, we need to analyze the magnetic fields produced by the two parallel wires carrying currents \( i_1 \) and \( i_2 \).

### Step 1: Understand the Magnetic Field Due to a Long Straight Current-Carrying Wire
The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2I}{r} \]
where \( \mu_0 \) is the permeability of free space.

### Step 2: Set Up the Problem
We have two long parallel wires:
- Wire 1 carries current \( i_1 \) (greater current).
- Wire 2 carries current \( i_2 \) (lesser current).

The distance from each wire to the midpoint (point P) is the same, let’s denote this distance as \( x \).

### Step 3: Calculate the Magnetic Field When Currents Are in the Same Direction
When both currents are in the same direction, the magnetic fields at point P due to both wires will have opposite directions. Thus, the net magnetic field \( B_{net1} \) at point P is given by:
\[ B_{net1} = B_1 - B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2i_1}{x} - \frac{\mu_0}{4\pi} \cdot \frac{2i_2}{x} \]
This simplifies to:
\[ B_{net1} = \frac{\mu_0}{4\pi x} \cdot 2(i_1 - i_2) \]
Given that \( B_{net1} = 10 \, \mu T \), we can write:
\[ \frac{\mu_0}{4\pi x} \cdot 2(i_1 - i_2) = 10 \times 10^{-6} \]
Let’s denote this as Equation (1).

### Step 4: Calculate the Magnetic Field When Currents Are in Opposite Directions
When the direction of \( i_2 \) is reversed, the magnetic fields at point P will add up:
\[ B_{net2} = B_1 + B_2 = \frac{\mu_0}{4\pi} \cdot \frac{2i_1}{x} + \frac{\mu_0}{4\pi} \cdot \frac{2i_2}{x} \]
This simplifies to:
\[ B_{net2} = \frac{\mu_0}{4\pi x} \cdot 2(i_1 + i_2) \]
Given that \( B_{net2} = 40 \, \mu T \), we can write:
\[ \frac{\mu_0}{4\pi x} \cdot 2(i_1 + i_2) = 40 \times 10^{-6} \]
Let’s denote this as Equation (2).

### Step 5: Divide Equation (2) by Equation (1)
Now, we can divide Equation (2) by Equation (1):
\[ \frac{\frac{\mu_0}{4\pi x} \cdot 2(i_1 + i_2)}{\frac{\mu_0}{4\pi x} \cdot 2(i_1 - i_2)} = \frac{40 \times 10^{-6}}{10 \times 10^{-6}} \]
This simplifies to:
\[ \frac{i_1 + i_2}{i_1 - i_2} = 4 \]

### Step 6: Cross Multiply and Rearrange
Cross multiplying gives us:
\[ i_1 + i_2 = 4(i_1 - i_2) \]
Expanding this, we have:
\[ i_1 + i_2 = 4i_1 - 4i_2 \]
Rearranging terms leads to:
\[ 3i_1 = 5i_2 \]

### Step 7: Find the Ratio \( \frac{i_1}{i_2} \)
From \( 3i_1 = 5i_2 \), we can express the ratio:
\[ \frac{i_1}{i_2} = \frac{5}{3} \]

### Final Answer
Thus, the ratio \( \frac{i_1}{i_2} \) is:
\[ \frac{i_1}{i_2} = \frac{5}{3} \]